zoukankan      html  css  js  c++  java
  • poj2262

                                                                                                   Goldbach's Conjecture
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 36877   Accepted: 14119

    Description

    In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:
    Every even number greater than 4 can be
    written as the sum of two odd prime numbers.

    For example:
    8 = 3 + 5. Both 3 and 5 are odd prime numbers.
    20 = 3 + 17 = 7 + 13.
    42 = 5 + 37 = 11 + 31 = 13 + 29 = 19 + 23.

    Today it is still unproven whether the conjecture is right. (Oh wait, I have the proof of course, but it is too long to write it on the margin of this page.)
    Anyway, your task is now to verify Goldbach's conjecture for all even numbers less than a million.

    Input

    The input will contain one or more test cases.
    Each test case consists of one even integer n with 6 <= n < 1000000.
    Input will be terminated by a value of 0 for n.

    Output

    For each test case, print one line of the form n = a + b, where a and b are odd primes. Numbers and operators should be separated by exactly one blank like in the sample output below. If there is more than one pair of odd primes adding up to n, choose the pair where the difference b - a is maximized. If there is no such pair, print a line saying "Goldbach's conjecture is wrong."

    Sample Input

    8
    20
    42
    0
    

    Sample Output

    8 = 3 + 5
    20 = 3 + 17
    42 = 5 + 37
    

    Source

    Ulm Local 1998
    思路:筛选法弄出素数,然后a从最小素数循环,遇到结果就直接输出。注意C语言比C++速度
    #include<stdio.h>
    #include<math.h>
    #include<string.h>
    #define MAXN 1000001
    bool vis[MAXN];
    
    int main()
    {
        //筛选法挑选素数
        memset(vis,0,sizeof(vis));
        for(int i = 2;i<(int)sqrt((double)MAXN);i++)
        {
            if(!vis[i])
            {
                for(int j=i*i ; j<MAXN;j+=i)
                    vis[j]=1;
            }
        }
        int n;
        while(scanf("%d",&n)!=EOF&&n!=0)
        {
                int a = 3;
                for(;a<n;a+=2)
                {
                    if(!vis[(n-a)]&&!vis[a])
                    {
                        printf("%d = %d + %d
    ",n,a,n-a);
                        break;
                    }
                }
        }
        return 0;
    }
    快些。
     
  • 相关阅读:
    2015的最后一天
    网络类型IPv4和IPv6什么意思?区别?
    2:文档编辑生成目录相关方法说明
    TCP与UDP的区别
    vs 2012 InstallShield Limited Edition Project 打包windows服务解析
    百科编辑器ueditor应用笔记
    百度编辑器Ueditor 初始化加载内容失败解决办法
    矩阵-DirectX与OpenGL的不同
    ios系统中各种设置项的url链接
    简单的优化处理 By LINQ TO SQL
  • 原文地址:https://www.cnblogs.com/jhldreams/p/3750646.html
Copyright © 2011-2022 走看看